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bmelyk

  • 3 years ago

I just wanna get my answer checked:

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  1. bmelyk
    • 3 years ago
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    The height at time t (s) of a weight, oscillating up and down at the end of a spring is given below. s(t) = 400 + 90sin(t)

  2. bmelyk
    • 3 years ago
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    a) Find the velocity at t = π/3 s. b) Find the acceleration at t = π/3 s

  3. bmelyk
    • 3 years ago
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    so for a) my equation i had was 90sin(t) and b) 90cos(t)

  4. bmelyk
    • 3 years ago
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    wondering if those are right.

  5. hartnn
    • 3 years ago
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    for velocity, u differentiate once, for acceleration, u differentiate twice.

  6. hartnn
    • 3 years ago
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    so NO

  7. hartnn
    • 3 years ago
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    first derivative of s(t) = 400 + 90sin(t) is ??

  8. bmelyk
    • 3 years ago
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    90cos(t) ??

  9. hartnn
    • 3 years ago
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    yes, thats your velocity

  10. hartnn
    • 3 years ago
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    and acceleration ?

  11. bmelyk
    • 3 years ago
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    -90sin(t)

  12. hartnn
    • 3 years ago
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    thats correct also :)

  13. hartnn
    • 3 years ago
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    now just put t values

  14. bmelyk
    • 3 years ago
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    so the answers are: A) 45 b) -45 sqrt3 ???

  15. hartnn
    • 3 years ago
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    yes.

  16. bmelyk
    • 3 years ago
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    woo hoooo. : )

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